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Leya [2.2K]
1 year ago
8

A line passes through the point(8,-5) and has a slope of " -3" Write an equation for this line.

Mathematics
2 answers:
shepuryov [24]1 year ago
3 0

Answer:

y=3x-29

Step-by-step explanation:

y=3*8*b

-5=3*8*b

b = 29

use slope-intercept formula

balandron [24]1 year ago
3 0

Answer:

y =-3x+19

Step-by-step explanation:

Let's refer to the formula for crafting linear equations:

y-x_{1} =m(x-x_{1} )

Now, substitute the given values in the formula and calculate:

y-x_{1} =m(x-x_{1} )\\\\y-(-5) =-3(x-(8))\\\\y+5 =-3x+24\\\\y =-3x+24-5\\\\y =-3x+19\\

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Rachel multiplied each side of x ≥ 2 by 3. She wrote the result as 3x ≤ 6. Explain the error Rachel made.
andrey2020 [161]

Answer:

Her mistake was to change the inequality sign from ≥ to ≤.

The sign only changes when you multiply or divide both sides of an inequality by a negative number.

6 0
2 years ago
Can someone help me with this math homework please!
Korvikt [17]

Answer:

Last answer.

Step-by-step explanation:

What I do is divide the terms by the previous terms, and if they all equal 1.5, that is your answer.

5 0
2 years ago
NEED ASAP
Gala2k [10]

Answer:

(- 2, 4 )

Step-by-step explanation:

Given endpoints (x₁, y₁ ) and (x₂, y₂ ) , then the midpoint is

( \frac{x_{1}+x_{2}  }{2} , \frac{y_{1}+y_{2}  }{2} )

Here (x₁, y₁ ) = A (4, 6 ) and (x₂, y₂ ) = B (- 8, 2 )

midpoint = ( \frac{4-8}{2} , \frac{6+2}{2} ) = ( \frac{-4}{2} , \frac{8}{2} ) = (- 2, 4 )

6 0
3 years ago
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If one rectangular solid with dimensions of 6 x 6 x10 millimeters is stacked on top of an identical rectangular solid so that th
ollegr [7]

Answer: 720

Step-by-step explanation: took the test

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Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -
GaryK [48]

Consider all parabolas:

1.

y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

2.

y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

3.

y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

4.

y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

5.

y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

6.

y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0
3 years ago
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